Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m;
vector<vector<int>> a;
struct TResult {
  vector<int> Rows, Cols;
  int GetSum() const {
    return accumulate(Rows.begin(), Rows.end(), 0) +
           accumulate(Cols.begin(), Cols.end(), 0);
  }
};
bool operator<(const TResult& lhs, const TResult& rhs) {
  return lhs.GetSum() < rhs.GetSum();
}
void Solve() {
  TResult bestResult;
  bool found = false;
  for (int cntCol = 0; cntCol <= a[0][0]; ++cntCol) {
    TResult result;
    result.Rows.resize(n);
    result.Cols.resize(m);
    bool ok = true;
    for (int i = 0; i < n; ++i) {
      result.Rows[i] = a[i][0] - cntCol;
      if (result.Rows[i] < 0) {
        ok = false;
      }
    }
    result.Cols[0] = cntCol;
    for (int i = 1; i < m; ++i) {
      result.Cols[i] = a[0][i] - result.Rows[0];
      if (result.Cols[i] < 0) {
        ok = false;
      }
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < m; ++j) {
        if (result.Rows[i] + result.Cols[j] != a[i][j]) {
          ok = false;
        }
      }
    }
    if (ok && (!found || result < bestResult)) {
      found = true;
      bestResult = result;
    }
  }
  if (!found) {
    cout << -1 << endl;
  } else {
    cout << bestResult.GetSum() << endl;
    const auto show = [](const vector<int>& values, const string prefix) {
      for (int i = 0; i < (int)values.size(); ++i) {
        for (int j = 0; j < values[i]; ++j) {
          cout << prefix << " " << i + 1 << endl;
        }
      }
    };
    show(bestResult.Rows, "row");
    show(bestResult.Cols, "col");
  }
}
bool Read() {
  if (!(cin >> n >> m)) {
    return false;
  }
  a.clear();
  a.resize(n);
  for (int i = 0; i < n; ++i) {
    a[i].resize(m);
    for (int j = 0; j < m; ++j) {
      cin >> a[i][j];
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  while (Read()) {
    Solve();
  }
  return 0;
}